I am interested in a proof of the following statement which seems intuitive, but is somehow really tricky:
Let $X$ be a stochastic process and let $(\mathcal{F}(t) : t \geq 0)$ be the filtration that it generates (unaugmented). Let $T$ be a bounded stopping time. Then we have
$\mathcal{F}(T) = \sigma(X(T \wedge t) : t \geq 0)$
I have a proof at hand (Bain and Crisan, Fundamentals of Stochastic Filtering, page 309), but in my opinion there is a major gap. I will try to explain the idea of proof.
Let $V$ be the space of functions $[0,\infty) \rightarrow \mathbb{R}$ equipped with the sigma algebra generated by the cylinder sets. Consider the canonical map $X^T:\Omega \rightarrow V$ which maps $\omega$ to the trajectory $t \mapsto X(t \wedge T(\omega),\omega)$. Then we have $\sigma(X(T \wedge t) : t \geq 0) = \sigma(X^T)$.
The difficult part is $\subseteq$. Let $A \in \mathcal{F}(T)$. We want to find a measurable map $g:V \rightarrow \mathbb{R}$ such that $1_A = g \circ X^T$, then we're done. It is now straightforward to show that $1_A$ is constant on sets where the sample paths of $X^T$ are constant. (To be more precise: for $\rho \in \Omega$ consider the set $\mathcal{M}(\rho) = \lbrace \omega : X(\omega,t) = X(\rho,t), 0 \leq t \leq T(\rho) \rbrace$. Then $T$ and $1_A$ are constant on every set of this form).
The problem is: this is not sufficient! It suffices to construct a map $g$ such that $1_A = g \circ X^T$, but how we can we know that $g$ is measurable? This is where the proof of Bain and Crisan comes up short IMO.
I can show this result only under the assumption that the map $X:\Omega \rightarrow V$ be surjective: Since $A \in \mathcal{F}(\infty)$, we have a measurable map $g$ such that $1_A = g \circ X$. Let $x \in V$. Then $T$ and $1_A$ are constant on the preimage of $x$ under $X$. Therefore, $g(x)$ does not depend on the values of $x$ after time $T$ (which is constant on the preimage of $x$). Since $X$ is surjective, we have $g(x) = g(K^Tx)$, where $K$ is the killing functional $K^tx(s) = x(t \wedge s)$. Hence, $g \circ X = g \circ X^T$, and we are done.
I think that this result could be a little bit deeper. I have seen two proofs of this for the special case that $X$ is the coordinate process on $C[0,\infty)$, one is given in the book of Karatzas & Shreve, Lemma 5.4.18. The fact that Karatzas proves this late in the book only in this special case somehow makes me think that the general case is not so easy.
I would really appreciate any comment or other reference for this result.
Best Answer
I've been worried about the same thing. Here's what I came up with:
Assume that $X$ is progressively measurable (e.g. cadlag), then the inclusion $\sigma (X_s^T : s \le t)\subset\mathcal{X}_T^0$ trivial. Without this condition, this inclusion won't hold in general, I believe. For the other direction:
$A \in \mathcal{X}_T^0 \Leftrightarrow \mathbb{1}_A$ is $\mathcal{X}_T^0$ measurable $\Leftrightarrow \mathbb{1}_A = Y_T$ for some $\mathcal{X}_t^0$-optional process $Y_t$.
Thus it suffices to show that for all optional processes $Y_t$, $Y_T$ is $\sigma (X_s^T: s \ge 0)$ measurable.
Now, the optional processes are generated by stochastic intervals of the form $[\sigma,\infty)$, so using a functional class argument it is enough to show that $\{\sigma \le T\} \in \sigma (X_s^T : s \ge 0)$ for all $\mathcal{X}_t^0$-stopping times $\sigma$, $T$. Do this via discretisation of $\sigma$, $T$ and taking limits. (Similarly we can prove that a $\mathcal{X}_t^0$-stopping time $T$ is a $\sigma (X_s^T : s \ge 0)$ stopping time, I believe.)
EDIT: I spoke too soon, this discretisation argument doesn't work unless the filtration is right continuous.... I have no idea how to proceed.
EDIT2: It appears that the proof is IV.100 of Probabilities and Potential (Dellacherie and Meyer)　though I do not currently have this to hand.